How do I prove there is no solution/solution to the single equation $150a+5b=54c$ provided $a,b,c$ must be different integer numbers ranging from 0-9?
I can verify it by running a computer program, but I am looking for mathematical reasoning.
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1b cant be 54 it must be between 0-9 as pre condition – Rejoy VM Apr 28 '20 at 14:38
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(0,0,0) cant be solution coz a,b,c must be unique as per precondition – Rejoy VM Apr 28 '20 at 14:43
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If you decompose by prime factors: $$150a+5b=54c\longrightarrow(2*3*5^2)a+5b=(2*3^3)c\longrightarrow 5*((2*3*5)a+b)=(2*3^3)c$$ So if that is correct, c must be 5, and we can divide the whole equation by 5 having $$(2*3*5)a + b = 2*3^3 \longrightarrow 30a+b=54$$ Now, we see that $a$ can't be equal to $0$ or $1$ because then $b$ would be grater than 9, absurd. On the other hand, if $a\geq2$, then $b$ would be negative, also absurd.
So we come to the conclusion that $150a+5b=54c$ has no solutions for $a,b,c$ integers between 0 and 9.
Alejandro Bergasa Alonso
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Clearly $c\equiv 0\mod 5$, so if there is a solution, we must have $c=5$, which means $a$ is either $0$ or $1$, clearly $a\neq 0$ as that would mean $b=54$, and if $a=1$, $b=24$, so there are no solutions in the given range
Loobear23
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